A 550 kg elevator starts from rest. It moves upward for 5.00 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?
W
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising = W


The formula for average power is avg. power = work/change in time
However, I am not sure how to find the work in the problem since no force or angle is given...Is gravity the only force doing work and is the angle 90?

Capuchin

the angle is 0 (you don't need to worry about it, because it's a 1d problem), the motor is doing work against the force of gravity.

tnhoots

Ok so the F in the problem is 9.8m/s^2. I would use cos0. However, the change in distance is unknown. Using the formula d=time * speed, I found it to be 8.75m. So, now would I use the work formula with the numbers (9.8m/s^2)(8.75m)(cos0). Which equals 85.57J. After I find that I would divide work by time (5secs) to find average power which is 17.15 W...just my thinking??

Capuchin

F = ma

your distance is wrong because it's accelerating, you need to use suvat equations

tnhoots

The force in this problem would be the mass of the elevator time the acceleration which is 1.75m/s correct. So f in the problem would be 962.5N. For distance I mult. the speed times the time. That is the only way I know to find distance.

Capuchin

acceleration is the acceleration of gravity!!!!

you need to use s = ut + 1/2at^2 to find distance traveled, because it is under constant acceleration.

I have to hit the sack now, but i will carry on tomorrow.

tnhoots

One more try:
f=mass*acceleration
f=550*9.8=5390N
change in distance=122.5m
cosign of angle 0degrees

w=f*change in d*cosign0
w=5390*122.5*cosign0
??? Correct??

Capuchin

Oh no, your acceleration for s = ut + 1/2 at^2 is 1.75/5

schmidtj88

The answer for the second part is an alternate formula for power which is velocity times force. So mass times gravity(9.8) times 1.75. That is the correct answer to the second part. I just got it right on my homework so its right.